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Multiplicative encoding of triangle formed by reading Pascal's triangle mod 2 (A047999).
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%I #84 Jan 21 2020 00:38:34

%S 2,6,10,210,22,858,1870,9699690,46,4002,7130,160660290,20746,

%T 1008940218,2569288370,32589158477190044730,118,21594,39530,

%U 3595293030,94754,17808161514,44788794490,7074421030108255253430,263258,141108130806,281595235990,296987147493893719182390,944729501606

%N Multiplicative encoding of triangle formed by reading Pascal's triangle mod 2 (A047999).

%C This is to A047999 "Triangle formed by reading Pascal's triangle mod 2" as A007188 "Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows." a(2^n - 1) = primorial(2^n) = A002110(A000079(n)). In row(n) the primes with exponent 1 form row(n) of a Sierpinski sieve, so this sequence is a kind of Gödelization of a Sierpinski sieve.

%C All terms are divisible by 2 and the n-th term, a(n-1), is also divisible by prime(n). This sequence appears as first column of the square array A255483; its second column A276804 is very similar, with all prime factors shifted to the net larger prime (cf. A003961). - _M. F. Hasler_, Sep 17 2016

%C a(n) is the n-th power of 6 in the ring defined in A329329. - _Peter Munn_, Jan 04 2020

%H Chai Wah Wu, <a href="/A123098/b123098.txt">Table of n, a(n) for n = 0..510</a>

%H C. Cobeli, A. Zaharescu, <a href="http://dx.doi.org/10.1080/10236198.2014.940337">A game with divisors and absolute differences of exponents</a>, Journal of Difference Equations and Applications, Vol. 20, #11 (2014) pp. 1489-1501, DOI: 10.1080/10236198.2014.940337. Also available as <a href="http://arxiv.org/abs/1411.1334">arXiv:1411.1334 [math.NT]</a>, 2014.

%F a(n) = Product_{i=0..n} p(i+1)^(C(n,i) mod 2).

%F a(n) = Product_{i=0..n} p(i+1)^T(n,i), where T(n,i) are as in A047999 and where Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).

%F From _Antti Karttunen_, Sep 18 2016: (Start)

%F a(n) = A007913(A007188(n)). [From the first comment.]

%F a(n) = A019565(A001317(n)).

%F (End)

%F a(0) = 2, and for n > 0, a(n) = A329329(a(n-1), 6). - _Peter Munn_, Jan 04 2020

%e a(0) = 2^T(0,0) = 2^1 = 2.

%e a(1) = 2^T(1,0) * 3^T(1,1) = 2^1 * 3^1 = 6.

%e a(2) = 2^T(2,0) * 3^T(2,1) * 5^T(2,2) = 2^1 * 3^0 * 5^1 = 10.

%e a(3) = 2^T(3,0) * 3^T(3,1) * 5^T(3,2) * 7^T(3,3) = 2^1 * 3^1 * 5^1 * 7^1 = 210.

%e a(4) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 = 22.

%e a(5) = 2^1 * 3^1 * 5^0 * 7^0 * 11^1 * 13^1 = 858.

%e a(6) = 2^1 * 3^0 * 5^1 * 7^0 * 11^1 * 13^0 * 17^1 = 1870.

%e a(7) = 2^1 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 * 19^1 = 9699690.

%e a(8) = 2^1 * 3^0 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 = 46.

%e a(9) = 2^1 * 3^1 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 = 4002.

%e a(10) = 2^1 * 3^0 * 5^1 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^1 = 7130.

%e a(11) = 2^1 * 3^1 * 5^1 * 7^1 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 * 31^1 * 37^1 = 160660290.

%e a(12) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^0 * 37^0 * 41^1 = 20746.

%e From _N. J. A. Sloane_, Feb 28 2015: (Start)

%e Factorizations of initial terms, from Cobeli-Zaharescu paper:

%e 2 = 2

%e 6 = 2*3

%e 10 = 2*5

%e 210 = 2*3*5*7

%e 22 = 2*11

%e 858 = 2*3*11*13

%e 1870 = 2*5*11*17

%e 9699690 = 2*3*5*7*11*13*17*19

%e 46 = 2*23

%e 4002 = 2*3*23*29

%e 7130 = 2*5*23*31

%e 160660290 = 2*3*5*7*23*29*31*37

%e 20746 = 2*11*23*41

%e 1008940218 = 2*3*11*13*23*29*41*43

%e 2569288370 = 2*5*11*17*23*31*41*47

%e 32589158477190044730 = 2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53

%e ... (End)

%e From _Jon E. Schoenfield_, Jun 09 2019: (Start)

%e n | Factorization of a(n)

%e ---+-----------------------------------------------

%e 0 | 2

%e 1 | 2* 3

%e 2 | 2 * 5

%e 3 | 2* 3* 5* 7

%e 4 | 2 *11

%e 5 | 2* 3 *11*13

%e 6 | 2 * 5 *11 *17

%e 7 | 2* 3* 5* 7*11*13*17*19

%e 8 | 2 *23

%e 9 | 2* 3 *23*29

%e 10 | 2 * 5 *23 *31

%e 11 | 2* 3* 5* 7 *23*29*31*37

%e 12 | 2 *11 *23 *41

%e 13 | 2* 3 *11*13 *23*29 *41*43

%e 14 | 2 * 5 *11 *17 *23 *31 *41 *47

%e 15 | 2* 3* 5* 7*11*13*17*19*23*29*31*37*41*43*47*53

%e ... (End)

%p f:=n->mul(ithprime(k+1)^(binomial(n,k) mod 2),k=0..n);

%p [seq(f(n),n=0..40)];

%t a[n_] := Product[Prime[k+1]^Mod[Binomial[n, k], 2], {k, 0, n}];

%t Table[a[n], {n, 0, 28}] (* _Jean-François Alcover_, Oct 01 2018, from Maple *)

%o (Python)

%o from operator import mul

%o from functools import reduce

%o from sympy import prime

%o def A123098(n):

%o return reduce(mul,(1 if ~(n-1) & k else prime(k+1) for k in range(n))) # _Chai Wah Wu_, Feb 08 2016

%o (Scheme) (define (A123098 n) (A019565 (A001317 n))) ;; _Antti Karttunen_, Sep 18 2016

%o (PARI) a(n) = prod (k=0, n, if (binomial(n,k)%2, prime(k+1), 1)) \\ _Rémy Sigrist_, Jun 09 2019

%Y Cf. A000040, A000120, A001316, A001317, A007188, A007318, A007913, A047999, A019565, A255484, A329329.

%Y First column of A255483.

%K nonn

%O 0,1

%A _Jonathan Vos Post_, Nov 05 2006

%E Further terms from _N. J. A. Sloane_, Feb 28 2015

%E Changed offset from 1 to 0, corresponding changes to formulas and examples from _Antti Karttunen_, Sep 18 2016